Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

int main(){
	int N;
	cin>>N;
	int i,j;
	vector<double>P(N+1);//Probabilities
	vector<vector<double>>DP(N+1,vector<double>(N+1,0));
	for(i=1;i<=N;i++){
		cin>>P[i];
	}
	DP[0][0]=1;
	for(i=1;i<=N;i++){
		DP[i][0]=1;
		for(j=1;j<=N;j++){
			DP[i][j]=(DP[i-1][j-1]*P[i])+(DP[i-1][j]*(1-P[i]));
		}
	}
	cout.precision(10);
	cout<<DP[N][(N+1)/2];
	return 0;
}